details

property	value
status	complete
benchmark	Ex4_4_Luc96b_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n063.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.65507793427 seconds
cpu usage	7.494804193
max memory	5.10246912E8

stage attributes

key	value
output-size	2727
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) NonTerminationLoopProof [COMPLETE, 0 ms] (4) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(g(X), Y) -> f(X, n__f(g(X), activate(Y))) f(X1, X2) -> n__f(X1, X2) activate(n__f(X1, X2)) -> f(X1, X2) activate(X) -> X Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: F(g(X), Y) -> F(X, n__f(g(X), activate(Y))) F(g(X), Y) -> ACTIVATE(Y) ACTIVATE(n__f(X1, X2)) -> F(X1, X2) The TRS R consists of the following rules: f(g(X), Y) -> f(X, n__f(g(X), activate(Y))) f(X1, X2) -> n__f(X1, X2) activate(n__f(X1, X2)) -> f(X1, X2) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = F(g(X'), n__f(g(X), X2)) evaluates to t =F(X, n__f(g(X), activate(X2))) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [X' / X'', X2 / activate(X2)] * Semiunifier: [X / g(X'')] -------------------------------------------------------------------------------- Rewriting sequence F(g(X'), n__f(g(g(X'')), X2)) -> ACTIVATE(n__f(g(g(X'')), X2)) with rule F(g(X'''), Y) -> ACTIVATE(Y) at position [] and matcher [X''' / X', Y / n__f(g(g(X'')), X2)] ACTIVATE(n__f(g(g(X'')), X2)) -> F(g(g(X'')), X2) with rule ACTIVATE(n__f(X1, X2')) -> F(X1, X2') at position [] and matcher [X1 / g(g(X'')), X2' / X2] F(g(g(X'')), X2) -> F(g(X''), n__f(g(g(X'')), activate(X2))) with rule F(g(X), Y) -> F(X, n__f(g(X), activate(Y))) Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (4) NO popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Standard